How can I calculate$$\int \frac {x\left(J_{ \frac 34}(\frac {x^2}{2})-J_{- \frac 54}(\frac {x^2}{2})\right)}{2J_{- \frac 14}(\frac {x^2}{2})}dx~?$$


1 Answer 1


You have some recurrence formulas that you can maybe use in this case: $$J_{n+1}(x)-J_{n-1}(x)=-2J'_n(x)$$ Since $n+1=\frac 3 4\implies n=-\frac 14$ And $$n-1=-\frac 5 4$$ So you have : $$J_{3/4}(x)-J_{-5/4}(x)=-2J'_{-1/4}(x)$$ And for the integral: $$ \begin{align} I=&\int \frac {x\left(J_{ \frac 34}(\frac {x^2}{2})-J_{- \frac 54}(\frac {x^2}{2})\right)}{2J_{- \frac 14}(\frac {x^2}{2})}dx \\ I=&-\int \frac x{J_{- \frac 14}(\frac {x^2}{2})} \frac {dJ_{-\frac 14}(x^2/2)}{d \frac {x^2}2}dx\\ I=&-\int \frac {dJ_{- \frac 14}(\frac {x^2}{2})} {J_{-\frac 14}(\frac {x^2}{2})}\\ I=&-\ln \left ( J_{- \frac 14}(\frac {x^2}{2}) \right ) \end{align} $$

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    $\begingroup$ That last sentence made me a little nauseous haha. Good work with the Bessel function identities. They're incredibly useful. $\endgroup$ Dec 18, 2019 at 17:24
  • $\begingroup$ @CameronWilliams Thanks. I made a mistake the integral can be evaluated nicely because the Bessel function is taken for $x^2/2$ so the derivative in the recurrence formula is nice. $\endgroup$ Dec 18, 2019 at 18:06
  • $\begingroup$ Indeed, very nice use of the identity! Don’t you want to complete the problem by evaluating the integral then? As you say, everything works out nicely... $\endgroup$ Dec 18, 2019 at 18:21
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    $\begingroup$ It is nice, i like it! $\endgroup$ Dec 18, 2019 at 18:41
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    $\begingroup$ Exactly, that’s the final answer to OP’s question. $\endgroup$ Dec 18, 2019 at 19:17

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